There are two different vats. The first vat has 20 gallons is being filled at a rate of 3 gallons per minute. The second vat has 108 gallons and is leaking fluid at the rate of 2 gallons per minute. After how many seconds will both vats have the same amount of liquid?

Question

Minutes :

Answer 1

Let's denote the time in minutes as \( t \).

1. For the first vat:

   • It starts with 20 gallons and is being filled at a rate of 3 gallons per minute.
   • The amount of liquid in the first vat after \( t \) minutes is given by: \[ \text{Amount in first vat} = 20 + 3t \]

2. For the second vat:

   • It starts with 108 gallons and is leaking at a rate of 2 gallons per minute.
   • The amount of liquid in the second vat after \( t \) minutes is given by: \[ \text{Amount in second vat} = 108 - 2t \]

To find when both vats have the same amount of liquid, we set the two equations equal to each other: \[ 20 + 3t = 108 - 2t \]

Now, let's solve for \( t \):

1. Add \( 2t \) to both sides: \[ 20 + 3t + 2t = 108 \] This simplifies to: \[ 20 + 5t = 108 \]

2. Subtract 20 from both sides: \[ 5t = 108 - 20 \] Which simplifies to: \[ 5t = 88 \]

3. Divide both sides by 5: \[ t = \frac{88}{5} = 17.6 \text{ minutes} \]

Now, to convert this into seconds: \[ t = 17.6 \times 60 = 1056 \text{ seconds} \]

Thus, both vats will have the same amount of liquid after 17.6 minutes or 1056 seconds.